M4_p4.3_z07.pdf

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ˇ RJESENJA

Zadatak 7.

Deriviraj funkcije: 1) f (x) = sin3 2x · cos2 3x ; 2) f (x) = 4 sin2 x − 8 sin x + 3 ; 3) f (x) = sin(x2 + 1) ; 4) f (x) = 2 sin x + 3 sin3 x ; 5) f (x) = sin2 (x3 ) ; 6) f (x) = 2x tg 2x ; √ 7) f (x) = sin 1 + x2 ; 8) f (x) = tg x2 · ctg x2 .

Rjeˇsenje.

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1)  f (x) = (sin3 2x · cos2 3x) = (sin3 2x) · cos2 3x + sin3 2x · (cos2 3x) == 3 sin2 2x cos 2x·2 cos2 3x+sin3 2x2 cos 3x(− sin 3x)·3 = 6 sin2 2x cos 3x(cos 2x cos 3x− sin 2x sin 3x) = 6 sin2 2x cos 3x cos 5x ; 2) f  (x) = (4 sin2 x− 8 sin x+ 3) = 8 sin x cos x− 8 cos x = 8 cos x(sin x− 1) ; 3)  f (x) = [sin(x2 + 1)] = 2x cos(x2 + 1) ; 4) f  (x) = (2 sin x + 3 sin3 x) = 2 cos x + 9 sin2 x cos x = cos x(2 + 9 sin2 x) ; 5) f  (x) = [sin2 (x3 )] = 2 sin(x3 ) cos(x3 ) · 3x2 = 3x2 sin(2x3 ) ; 4x 1 6) f  (x) = (2x tg 2x) = 2 tg 2x + 2x · · 2 = 2 tg 2x + ; 2 cos 2x cos2 2x √ √ √ 1 x 7) f  (x) = (sin 1 + x2 ) = cos 1 + x2 · √ ·2x = √ cos 1 + x2 ; 2 1 + x2 1 + x2 1 8) f  (x) = (tg x2 · ctg x2 ) = (tg x2 ) ctg x2 + tg x2 · (ctg x2 ) = · 2 x2 cos   1 2x 2x − = 0. 2x · ctg x2 + tg x2 − 2 2 · 2x = sin x sin x2 cos x2 sin x2 cos x2